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(cos A + cos b)^2 + (sin A + sin B)^2=?
@apoorvk @satellite73 @A.Avinash_Goutham @ajprincess @AndrewNJ
okay, expand the two whole-squares, it's simple enough...
the options are 4cos^2(a-b/2) 4 sin a/2 cos b/2
@apoorvk after expanding i got this then next..............
great, now.. cosAcosB + sinAsinB = cos(A-B) Use this^
ugh, trig identities are a drag. Even after deriving them myself I simply CANNOT seem to memorize them
spread it so we have (cosa)^2+(cosb)^2+2(cosa)(cosb)+(sina)^2+(sinb)^2+2(sina)(sinb) =2+2cos(a-b)
(cosa)^2+(sina)^2=1 (cosb)^2+(sinb)^2=1 (cosa)(cosb)+(sina)(sinb)=cos(a-b)
ok @apoorvk so it is not over yet!!@
Continuing from where @AndrewNJ left, \[2(1+\cos(a-b))\] now we know that \[1 + \cos 2a = 2 \cos^2 a\] Now try :D
@shivam_bhalla did nt understand!
= \[2(1+\cos 2\frac{(a-b)}{2})\] Now we have a identity, \[1 + \cos 2x = 2 \cos^2 x\] in your case, \[x = \frac{a-b}{2}\] Now try
i got it thxx @shivam_bhalla
@shivam_bhalla plzz help http://openstudy.com/study?signup#/updates/4fd60c0be4b04bec7f170b7d